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 A006566 Dodecahedral numbers: a(n) = n*(3*n - 1)*(3*n - 2)/2. (Formerly M5089) 30
 0, 1, 20, 84, 220, 455, 816, 1330, 2024, 2925, 4060, 5456, 7140, 9139, 11480, 14190, 17296, 20825, 24804, 29260, 34220, 39711, 45760, 52394, 59640, 67525, 76076, 85320, 95284, 105995, 117480, 129766, 142880, 156849, 171700, 187460, 204156 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Schlaefli symbol for this polyhedron: {5,3} A093485 = first differences; A124388 = second differences; third differences = 27. - Reinhard Zumkeller, Oct 30 2006 One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010 From Peter Bala, Sep 09 2013: (Start) a(n) = binomial(3*n,3). Two related sequences are binomial(3*n+1,3) (A228887) and binomial(3*n+2,3) (A228888). The o.g.f.'s for these three sequences are rational functions whose numerator polynomials are obtained from the fourth row [1, 4, 10, 16, 19, 16, 10, 4, 1] of the triangle of trinomial coefficients A027907 by taking every third term: Sum_{n >= 1} binomial(3*n,3)*x^n = (x + 16*x^2 + 10*x^3)/(1-x)^4; Sum_{n >= 1} binomial(3*n+1,3)*x^n = (4*x + 19*x^2 + 4*x^3)/(1-x)^4; Sum_{n >= 1} binomial(3*n+2,3)*x^n = (10*x + 16*x^2 + x^3)/(1-x)^4. (End) REFERENCES N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS T. D. Noe, Table of n, a(n) for n = 0..1000 Tanya Khovanova, Recursive Sequences Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75. Victor Meally, Letter to N. J. A. Sloane, no date. Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992. Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992. Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1). FORMULA G.f.: x(1 + 16x + 10x^2)/(1 - x)^4. a(n) = A000292(3n-3) = A054776(n)/6 = n*A060544(n). a(n) = C(n+2,3) + 16 C(n+1,3) + 10 C(n,3). a(0)=0, a(1)=1, a(2)=20, a(3)=84, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Jul 24 2013 a(n) = binomial(3*n,3). a(-n) = - A228888(n). Sum_{n>=1} 1/a(n) = 1/2*( sqrt(3)*Pi - 3*log(3) ). Sum_{n>=1} (-1)^n/a(n) = 1/3*sqrt(3)*Pi - 4*log(2). - Peter Bala, Sep 09 2013 a(n) = A006564(n) + A035006(n). - Peter M. Chema, May 04 2016 E.g.f.: x*(2 + 18*x + 9*x^2)*exp(x)/2. - Ilya Gutkovskiy, May 04 2016 MAPLE A006566:=(1+16*z+10*z**2)/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation MATHEMATICA Table[n(3n-1)(3n-2)/2, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2011 *) LinearRecurrence[{4, -6, 4, -1}, {0, 1, 20, 84}, 40] (* Harvey P. Dale, Jul 24 2013 *) CoefficientList[Series[x (1 + 16 x + 10 x^2)/(1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 11 2015 *) PROG (PARI) a(n)=n*(3*n-1)*(3*n-2)/2 (Haskell) a006566 n = n * (3 * n - 1) * (3 * n - 2) `div` 2 a006566_list = scanl (+) 0 a093485_list  -- Reinhard Zumkeller, Jun 16 2013 (MAGMA) [n*(3*n-1)*(3*n-2)/2: n in [0..40]]; // Vincenzo Librandi, Dec 11 2015 CROSSREFS Cf. A027907, A228887, A228888. Cf. A000292 (tetrahedral numbers), A000578 (cubes), A005900 (octahedral numbers), A006564 (icosahedral numbers). Sequence in context: A044207 A044588 A172221 * A205312 A268888 A211158 Adjacent sequences:  A006563 A006564 A006565 * A006567 A006568 A006569 KEYWORD nonn,easy,nice AUTHOR EXTENSIONS More terms from Henry Bottomley, Nov 23 2001 STATUS approved

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